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xxMikexx [17]
3 years ago
10

Identify the spectator ions in this reaction. Check all that apply. 2H+ + CrO4– + Ba2+ + 2OH– Ba2+ + CrO4– + 2H2O H+ CrO4– Ba2+

OH– 2H2O
Chemistry
2 answers:
Luba_88 [7]3 years ago
6 0

Answer: B and C on the tenth question of the assignment

bagirrra123 [75]3 years ago
5 0
Answer: CrO₄⁻ and  Ba²⁺


Explanation:


1) Chemical equation given:


2H⁺ + CrO₄⁻ + Ba²⁺ + 2OH⁻ → Ba²⁺ + CrO₄⁻ + 2H₂O


2) Analysis


That is an oxidation-reduction equation (some species are been oxidized and others are being reduced).


The given equation is known as total ionic equation, because it shows all the species as ions that are part of the reaction.


2) Specator ions


Spectator ions are the ions that do not change their oxidation state and are easily identified as they are the same in the reactant and product sides.


Here the ions that are the same in the reactant and product sides are:

CrO₄⁻ and  Ba²⁺


3) Addtitional explanation.


Once you identify the spectator ions you can delete them from the equation to obtain the net ionic equation , which in this case turns to be:


2H⁺ + 2OH⁻ →  2H₂O


But this is  not part of the question; it is some context to help you understand the use of the spectator ions concept.
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Number of grams of hydrogen than can be prepared from 6.80g of aluminum​
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Answer:

0.7561 g.

Explanation:

  • The hydrogen than can be prepared from Al according to the balanced equation:

<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

  • Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

<em>Using cross multiplication:</em>

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

  • Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

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